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Let $S^2=\{(x,y,z)\in \mathbb{R}^3|x^2+y^2+z^2=1\}$ and $S^1=\{(s,t)\in \mathbb{R}^2|s^2+t^2=1\}$. Suppose that $\mathbb{Z}/2\mathbb{Z}$ acts on $S^2\times S^1$ in such a way that the generator of $\mathbb{Z}/2\mathbb{Z}$ maps $$ (x,y,z)\mapsto(-x,-y,-z) \ \ \ and \ \ \ (s,t)\mapsto (-s,t) $$ (Note that $t$ is invariant). I would like to know what the free quotient space $(S^2\times S^1)/\mathbb{Z}/2\mathbb{Z}$ looks like. Since the action preserves orientation of $S^2\times S^1$, it should be oriented 3-manifold.

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$S^2/\mathbb{Z}_2 \simeq \mathbb{R}P^2$ this is the definition of projective space.

$S^1/\mathbb{Z}_2 \simeq [0,1]$. You fold the circle, it is homeomorphic to a line segment.

Your resulting space is $(S^2 \times S^1)/\mathbb{Z}_2 \simeq \mathbb{R}P^2 \times [0,1]$.

This 3-manifold is not oriented since $\vec{x} \mapsto -\vec{x}$ preserves orientation of $S^2$.

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Why do you assume that $(A\times B)/C \cong (A/C)\times(B/C)$? – Steven Stadnicki Oct 11 '12 at 0:10
(Specifically (since my earlier comment expired), the original question makes it clear that the two actions aren't acting independently; the poster wants their spaced 'chopped in half', not chopped in quarters.) – Steven Stadnicki Oct 11 '12 at 0:17
Thanks for the answer and comments. I don't think the answer is not the product of the quotients. I think that the quotient 3-fold should be oriented without boundary. – M. K. Oct 11 '12 at 1:47

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